\(\displaystyle f(x)\) is a continous function at \(\displaystyle R\) which gets a local maximum at point \(\displaystyle x_{0} \). i need to prove - formal proof, not just words - that if\(\displaystyle f\) doens't have any other extremas, f gets maximum at \(\displaystyle x_{0} \).
question in Weierstrass Theorem
- Thread starter orir
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"\(\displaystyle f(x)\) is an extrema value if \(\displaystyle f(x)\) is a local minimum or a local maximum of \(\displaystyle f\). in this case we'd say that the point \(\displaystyle (x,f(x)) \) on \(\displaystyle f\)'s graph is called an extrema of \(\displaystyle f\)."
Well, in particular, does your definition of local minimum or local maximum imply differentiability at those points? For example f(x) = |x^2-1| has only one point in which f'(x)=0, and it is at x=0. The point (0,1) is the only local maximum but certainly this function has no maximum.
i don't know why, but i in my course they assume this is true... so maybe we need to ignore some other possibilities... i don't know..
i don't know why, but i in my course they assume this is true... so maybe we need to ignore some other possibilities... i don't know..
If you assume f(x) is differentiable, then this is true. Since there is a local maximum there, we must have for some \(\displaystyle \delta >0\) that \(\displaystyle f'(x) > 0\) for \(\displaystyle x\in (x_0-\delta, x_0)\) and \(\displaystyle f'(x) < 0\) for \(\displaystyle x\in (x_0,x_0+\delta)\).
But since \(\displaystyle f(x)\) has no other extrema, that means \(\displaystyle f'(x)\) cannot change sign more than once. That means \(\displaystyle \delta\) can be chosen arbitrarily large (though there may be two-dimensional "saddle points"). In fact \(\displaystyle f'(x)\ge 0\) for \(\displaystyle x<x_0\) and \(\displaystyle f'(x) \le 0\) for \(\displaystyle x>x_0\).
Can you finish this and make it "formal"?
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